Freshers Interviews: Microsoft Written Test Questions

1. There are 2 sorted arrays A and B of size n each. Write an algorithm to find the median of the array obtained after merging the above 2 arrays(i.e. array of length 2n). Full points are given for the solution bearing efficiency O(log n). Partial points for the efficiency O(n).
2. Given a string *Str of ASCII characters, write the pseudo code to remove the duplicate elements present in them. For example, if the given string is "Potato", then, the output has to be "Pota". Additional constraint is, the algorithm has to be in-place( no extra data structures allowed) . Extend your algorithm to remove duplicates in the string which consisted of UNICODE characters.
3. Given that, there are 2 linked lists L1 and L2. The algorithm finds the intersection of the two linked lists. i.e' "L1 intersection L2 " ( similar to intersection of 2 sets ). Write all the possible test cases involved in the above given problem.

Please, post the answer if you know them.

double findMedianLog(int * a, int * b, int size) {
int m1 = a[size / 2];
int m2 = b[size / 2];

if (size == 2) {
return (max(a[0], b[0]) + min(a[1], b[1])) / 2;
}
if (m1 < m2) {
return findMedianLog(a + size / 2, b, size - size / 2);
}
return findMedianLog(b + size / 2, a, size - size / 2);
}

For 1st qns think on these lines for a logn algorithm:
compare the medians of both arrays (in O(1) since it is sorted). The problem size will be reduced to n elements by this single comparison. Then continue

main()
{
char *str = "abababababcdefefefefefegyuiopl";
char *p = str;
char *q = p + 1;
char *check = str;

while (*p != '\0')
{
if (*q == '\0')
{
p++;
q = p + 1;
}
while (( *q != *p ) && (*q != '\0'))
q++;

if (*p == *q)
{
check = q + 1;
if (*check == '\0')
*q ='\0';
strcpy(q,check);
q = p + 1;
}
}

printf ("\n%s",str);
}

For Question 2:

* Arrays
* Articles
* Bit Magic
* C/C++ Puzzles
* GFacts
* Linked Lists
* MCQ
* Misc
* Output
* Strings
* Trees

Print all the duplicates in the input string.
March 22, 2009

Write an efficient C program to print all the duplicates and their counts in the input string

Algorithm: Let input string be “geeksforgeeks”
1: Construct character count array from the input string.

count['e'] = 4
count['g'] = 2
count['k'] = 2
……

2: Print all the indexes from the constructed array which have value greater than 0.

Solution
view source
print?
# include
# include
# define NO_OF_CHARS 256

/* Returns an array of size 256 containg count
of characters in the passed char array */
int *getCharCountArray(char *str)
{
int *count = (int *)calloc(sizeof(int), NO_OF_CHARS);
int i;

for (i = 0; *(str+i); i++)
count[*(str+i)]++;

return count;
}

/* Print duplicates present in the passed string */
void printDups(char *str)
{
int *count = getCharCountArray(str);
int i;
char temp;

for (i = 0; i < NO_OF_CHARS; i++)
if(count[i] > 1)
printf("%c, count = %d \n", i, count[i]);
}

/* Driver program to test to pront printDups*/
int main()
{
char str[] = "test string";
printDups(str);
getchar();
return 0;

}
NOTE: Copied from geeksforgeeks.org site....

char *remove_duplicates(char *str)
{
char *str1, *str2;

if(!str)
return str;

str1 = str2 = str;

while(*str2)
{
if(strchr(str, *str2)<str2)
{
str2++;
continue;
}

*str1++ = *str2++;
}
*str1 = '\0';

return str;
}

Can anyone please give the solution for the last problem.

char * remove_dup_chars(char *str)
{
int len=strlen(str);
char *tmp=str,*tmp2=str;
int hole=0;
while(tmp && *tmp!='\0') {
hole=0;
tmp2=(tmp+1);
while(tmp2 && *tmp2!='\0') {
if(!hole && *tmp2==*tmp) {
hole=1;
}
if(hole) {
while(*(tmp2+hole) == *tmp) {
hole++;
}
*tmp2=*(tmp2+hole);
}
tmp2++;
}
tmp++;
}
return str;
}


int main()
{
char a[]="abcdabedaaabcf";
printf("orig:%s\n",a);
printf("after rem dup:%s\n",remove_dup_chars(a));
memcpy(a,"Potato",strlen("Potato")+1);
printf("orig:%s\n",a);
printf("after rem dup:%s\n",remove_dup_chars(a));

return 0;
}

int main()

{

char str[]="ababdfdfhjiab";
char *ptr;
int i;
printf("\n %s ",str);
for(i=0;i<strlen(str)-1;i++)
{
for(ptr=&str[i+1];*ptr!='\0';ptr++)
{
if(*ptr==str[i])
strcpy(ptr,ptr+1);
}
}
printf("\n %s ",str);
}

for the first question the method is very easy actually:

we can do it using a recursive function:
find the median of both the arrays:
suppose m1 is the median of the first array
m2 is the median of the second array

then if m1 > m2 then again call that recursive function in the right part of the second array and in the left part of the first array.

continue this step until the number of elements in the array is not two

when the number of elements in both the arrays are two then do the following:

suppose in the first array the elements are :

f1 f2

and in the second the elements are

s1 s2

then if

f1 < s2
return (f2+s1)/2

else
return (f1+s2)/2

#include
#include
#include
main(){
char s[100] = "aabcdeefgghijjkllllmnooo";
int l = strlen(s);
int i;
int count[226] = {0};
for(i=0;i1){
count[s[i]] = -1;
i++;
}
else if(count[s[i]]==-1){
int j;
for(j=i+1;j<l;j++){
s[j-1] = s[j];
}
l--;
}
else
i++;
}
s[l] = '\0';
printf("%s\n",s);
system("pause");
}

visit geeks for geeks.org

abe yaad kaise nahi rehta be gadhe aadhe question post kerta hai........... bhak

sorry be faaltu me maarli....galat samjha tha tujhe tu bahut achha insaan haibe....

IN 2nd question it is given no extra data structure should be used (IN place).so can we use extra 256 size array as hash table

Hi

I read this post two times.

I like it so much, please try to keep posting.

Let me introduce other material that may be good for our community.

Source: Computer programmer interview questions

Best regards
Henry

// code for question 1 : median.c

// Prem Raj... IITJ

#include
#include

void median(int* p1,int* p2,int* p3,int* p4);

int a[5] = {1,2,3,5,6};
int b[5] = {6,7,8,9,10};

int main()
{


int i=0,j=4,k=0,l=4;

median(&i,&j,&k,&l);

printf("Median is %d and %d\n",a[i],b[k]);

return 0;

}

void median(int* i,int* j,int* k,int* l){

int n1,n2,x;

if(*i > *j || *k > *l)
{
printf("Invalid array\n");
exit(0);
}

if(*i == *j)
{
return;
}

if(*i == *j-1)
{
if(a[*i]>b[*k])
{
if(a[*j]b[*l])
a[*j] = b[*l];
*i = *j;
*l = *k;
}
}

n1 = (*i+*j)/2;
n2 = (*k+*l)/2;
x = (*j-*i+1)%2;

if(a[n1] > b[n2])
{

if(x==0)
{*j = n1+1;
*k = n2;}
else
{*j = n1;
*k=n2;}
}

else
{

if(x==0)
{*l=n2+1;
*i=n1;}
else
{*l=n2;
*i=n1;}
}

median(i,j,k,l);
}